$\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ for distinct prime $p,q$

Solution 1:

Clearly $\mathbb Q(\sqrt p\sqrt[3]q)\subseteq \mathbb Q(\sqrt p,\sqrt[3] q)$. Also, if $\alpha = \sqrt p\sqrt[3]q$, then we see that $\alpha^3=pq\sqrt p$ and also $\frac{\alpha^3}{pq}=\sqrt p\in \mathbb Q(\sqrt p\sqrt[3]q)$. Similarly, $\frac{\alpha^4}{p^2q}=\sqrt[3]q\in\mathbb Q(\sqrt p\sqrt[3]q)$. We conclude $\mathbb Q(\sqrt p\sqrt[3]q)\supseteq \mathbb Q(\sqrt p,\sqrt[3] q)$.

For the degree of the extension, note that clearly $[\mathbb Q(\sqrt p):\mathbb Q]=2$ and $[\mathbb Q(\sqrt[3] q):\mathbb Q]=3$, hence $[\mathbb Q(\sqrt p,\sqrt[3] q):\mathbb Q]$ is

• either 2 or 6 from the tower $\mathbb Q(\sqrt p,\sqrt[3] q)\rightarrow \mathbb Q(\sqrt p)\rightarrow \mathbb Q$
• either 3 or 6 from the tower $\mathbb Q(\sqrt p,\sqrt[3] q)\rightarrow \mathbb Q(\sqrt[3] q)\rightarrow \mathbb Q$

Hence $[\mathbb Q(\sqrt p,\sqrt[3] q):\mathbb Q]=6$ and we conclude that the minimal polynomial of $\sqrt p\sqrt[3]q$ is of degree $6$, hence it is the obvious candidate $X^6-p^3q^2$.

Bonus questions: Where did I use that $p,q$ are prime? Where did I use that $p\ne q$?